\documentclass[a4paper,11pt]{article}

\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{lmodern}
\usepackage[french]{babel}
\usepackage[a4paper,top=3.2cm,bottom=2cm,left=2cm,right=2cm]{geometry}
\usepackage{graphicx}
\usepackage[export]{adjustbox}
\usepackage{fancyhdr}
\usepackage{qrcode}
\usepackage[table]{xcolor}
\usepackage{amsmath,amssymb}
\usepackage{enumitem}
% Puces nettes (babel-french remplace sinon la puce par un tiret « – » peu pro).
% Vaut pour TOUTES les fiches qui importent ce module. AtBeginDocument pour
% passer après la personnalisation des listes faite par babel-french.
\AtBeginDocument{%
  \setlist[itemize]{leftmargin=1.8em,itemsep=0.5ex,topsep=0.4ex,parsep=0pt}%
  \setlist[itemize,1]{label=\textcolor{red!75!black}{\small\textbullet}}%
  \setlist[itemize,2]{label=\textcolor{red!75!black}{\scriptsize\textbullet}}%
}
% Titres compacts (gain de place vertical, fiches plus denses) — vaut pour tous.
\usepackage{titlesec}
\titlespacing*{\section}{0pt}{1.4ex}{0.6ex}
\titlespacing*{\subsection}{0pt}{1.0ex}{0.3ex}
% multicol : listes denses (calcul mental) sur 2 colonnes — neutre tant qu'il
% n'est pas utilisé, dispo pour tous les générateurs.
\usepackage{multicol}
\setlength{\columnsep}{1.5em}
% needspace : empêche un titre d'exercice de rester orphelin en bas de page
% (réserve quelques lignes ; sinon le titre passe à la page suivante avec sa
% liste). Neutre tant qu'il n'est pas appelé — dispo pour tous les générateurs.
\usepackage{needspace}
\usepackage{tikz}
\usetikzlibrary{decorations.text}

\usepackage{tgheros}
\pagestyle{fancy}
\fancyhf{}
\setlength{\headheight}{75pt}
\setlength{\headsep}{10pt}

\newcommand{\logowouf}{%
\begin{tikzpicture}[baseline={(current bounding box.center)},scale=0.55,
                    line cap=round,line join=round]
    \path[use as bounding box] (-2.20,-1.45) rectangle (2.20,2.20);
    \path[
        postaction={decorate},
        decoration={
            text along path,
            text={|\fontfamily{qhv}\bfseries\fontsize{9}{9}\selectfont|site2wouf.fr},
            text align={center},
            raise=0.03ex
        }
    ] (-1.29,0.48) arc[start angle=160,end angle=20,radius=1.29];
    \begin{scope}[xshift=-0.06cm]
        \fill[black] (-0.66,0.14) ellipse (0.21 and 0.31);
        \fill[black] ( 0.66,0.14) ellipse (0.21 and 0.31);
        \fill[black] (-0.23,0.49) ellipse (0.22 and 0.35);
        \fill[black] ( 0.23,0.49) ellipse (0.22 and 0.35);
        \fill[black]
            (0,-0.10)
            .. controls (-0.36,-0.02) and (-0.63,-0.24) .. (-0.58,-0.58)
            .. controls (-0.52,-0.93) and (-0.16,-1.12) .. (0,-0.90)
            .. controls (0.16,-1.12) and (0.52,-0.93) .. (0.58,-0.58)
            .. controls (0.63,-0.24) and (0.36,-0.02) .. (0,-0.10)
            -- cycle;
    \end{scope}
\end{tikzpicture}%
}

\newcommand{\titre}{Trigonométrie}
\newcommand{\soustitre}{Cycle 4 --- Fiche n\textdegree 6}

\fancyhead[C]{%
\raisebox{4pt}{%
\begin{minipage}[c]{0.15\textwidth}
    \centering
    \logowouf
\end{minipage}
\begin{minipage}[c]{0.62\textwidth}
    \centering
    {\Large\bfseries \titre\par}
    {\small\color{gray} \soustitre}
\end{minipage}
\begin{minipage}[c]{0.21\textwidth}
    \centering
    \qrcode[height=1.5cm]{https://site2wouf.fr/exercices_trigonometrie.php?x=6}\\[2pt]
    {\small\ttfamily\bfseries\color{red} TRIG0006}
\end{minipage}%
}
}

\begin{document}

\sloppy

\needspace{4\baselineskip}
\section*{
Exercice 1}

 Dans le triangle FBN rectangle en F, on sait que :
\begin{itemize}
\item FB=3,1 cm
\item BN=9,4 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{FNB}\).

\needspace{4\baselineskip}
\section*{
Exercice 2}

 Dans le triangle PGB rectangle en P, on sait que :
\begin{itemize}
\item PG=2,2 cm
\item \(\widehat{PGB}\) = 50\textdegree{}
\end{itemize}

Après avoir fait un schéma, calcule la longueur du segment [BG]. (Arrondir au dixième)

\needspace{4\baselineskip}
\section*{
Exercice 3}

 Dans le triangle NHD rectangle en N, on sait que :
\begin{itemize}
\item NH=2,8 cm
\item HD=7,2 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{NHD}\).
\needspace{4\baselineskip}
\section*{
Exercice 4}

 Dans le triangle JDV rectangle en J, on sait que :
\begin{itemize}
\item DV=4,8 cm
\item \(\widehat{DVJ}\) =  30\textdegree{}
\end{itemize}

Après avoir fait un schéma, calcule la longueur du segment [JV]. (Arrondir au dixième)

\needspace{4\baselineskip}
\section*{
Exercice 5}

 Dans le triangle STD rectangle en S, on sait que :
\begin{itemize}
\item TD=1,1 cm
\item \(\widehat{STD}\) = 65\textdegree{}
\end{itemize}

Après avoir fait un schéma, calcule la longueur du segment [ST]. (Arrondir au dixième)

\newpage
\section*{\textcolor{red}{Correction}}

\needspace{4\baselineskip}
\subsection*{
Exercice 1}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,2.856) -- (1.037,2.643) -- (1.052,2.430) -- (1.045,2.212) -- (1.045,1.998) -- (1.061,1.789) -- (1.039,1.570) -- (1.039,1.358) -- (1.050,1.144);
  \draw[black] (1.050,1.144) -- (1.398,1.135) -- (1.753,1.142) -- (2.100,1.134) -- (2.451,1.133) -- (2.800,1.151) -- (3.147,1.148) -- (3.501,1.144) -- (3.851,1.152) -- (4.201,1.150) -- (4.548,1.147) -- (4.902,1.142) -- (5.250,1.134) -- (5.598,1.152) -- (5.950,1.144);
  \draw[black] (5.950,1.144) -- (5.624,1.262) -- (5.297,1.377) -- (4.969,1.483) -- (4.640,1.596) -- (4.319,1.714) -- (3.988,1.832) -- (3.660,1.933) -- (3.342,2.064) -- (3.009,2.175) -- (2.681,2.285) -- (2.358,2.395) -- (2.031,2.511) -- (1.703,2.637) -- (1.373,2.734) -- (1.050,2.856);
  \draw[black] (1.050,1.315) -- (1.051,1.258) -- (1.044,1.203) -- (1.050,1.144);
  \draw[black] (1.050,1.144) -- (1.106,1.144) -- (1.163,1.140) -- (1.221,1.144);
  \draw[black] (1.050,1.315) -- (1.106,1.315) -- (1.166,1.315) -- (1.221,1.315);
  \draw[black] (1.221,1.315) -- (1.215,1.259) -- (1.221,1.201) -- (1.221,1.144);
  \draw[black] (5.569,1.277) -- (5.569,1.265) -- (5.561,1.256) -- (5.563,1.244) -- (5.557,1.234) -- (5.558,1.222) -- (5.558,1.211) -- (5.561,1.199) -- (5.544,1.190) -- (5.549,1.178) -- (5.547,1.167) -- (5.555,1.155) -- (5.547,1.144);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,1.011) {F};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,2.923) {B};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,1.011) {N};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {3,1 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {9,4 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.293,1.189) {?};
\end{tikzpicture}\end{center}


 Dans le triangle FBN rectangle en F, on cherche une relation entre l'angle aigu   \(\widehat{FNB}\) son coté opposé et l'hypoténuse du triangle.

\par\addvspace{.5em}
sin(\(\widehat{FNB}\)) = \(\dfrac{FB}{BN}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{FNB}\)) =\(\dfrac{3{,}1}{9{,}4}\)
\par\addvspace{.5em}
 On a donc \(\widehat{FNB}\) = ArcSin(\(\dfrac{3{,}1}{9{,}4}\)) \(\approx\) 19\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.832,3.400) -- (1.841,3.049) -- (1.842,2.703) -- (1.828,2.350) -- (1.821,1.999) -- (1.826,1.652) -- (1.823,1.303) -- (1.828,0.947) -- (1.832,0.600);
  \draw[black] (1.832,0.600) -- (2.168,0.589) -- (2.500,0.601) -- (2.830,0.592) -- (3.169,0.588) -- (3.502,0.602) -- (3.832,0.601) -- (4.164,0.594) -- (4.504,0.592) -- (4.838,0.606) -- (5.168,0.600);
  \draw[black] (5.168,0.600) -- (4.903,0.808) -- (4.649,1.028) -- (4.395,1.247) -- (4.136,1.450) -- (3.890,1.684) -- (3.631,1.891) -- (3.367,2.100) -- (3.105,2.316) -- (2.861,2.546) -- (2.606,2.757) -- (2.336,2.962) -- (2.081,3.177) -- (1.832,3.400);
  \draw[black] (1.832,0.880) -- (1.832,0.787) -- (1.837,0.695) -- (1.832,0.600);
  \draw[black] (1.832,0.600) -- (1.923,0.600) -- (2.017,0.600) -- (2.112,0.600);
  \draw[black] (1.832,0.880) -- (1.924,0.883) -- (2.018,0.879) -- (2.112,0.880);
  \draw[black] (2.112,0.880) -- (2.115,0.788) -- (2.111,0.692) -- (2.112,0.600);
  \draw[black] (2.137,3.144) -- (2.114,3.126) -- (2.095,3.104) -- (2.072,3.087) -- (2.052,3.065) -- (2.029,3.046) -- (2.001,3.036) -- (1.972,3.032) -- (1.945,3.020) -- (1.918,3.012) -- (1.889,3.005) -- (1.860,3.003) -- (1.832,2.990);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.565,0.467) {P};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.565,3.467) {G};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.268,0.467) {B};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.332,1.933) {2,2 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.113,2.729) {50°};
\end{tikzpicture}\end{center}


 Dans le triangle PGB rectangle en P, on cherche une relation entre l'angle aigu   \(\widehat{PGB}\) son coté adjacent et l'hypoténuse du triangle.

cos(\(\widehat{PGB}\)) = \(\dfrac{PG}{GB}\)
\par\addvspace{.5em}
 d'où cos(50\textdegree{}) = \(\dfrac{2{,}2}{GB}\)
\par\addvspace{.5em}
On a donc GB = \(\dfrac{2{,}2}{\cos(50^{\circ})}\)\(\approx\) 3,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,3.034) -- (1.061,2.775) -- (1.062,2.514) -- (1.052,2.255) -- (1.045,2.002) -- (1.045,1.745) -- (1.041,1.480) -- (1.052,1.223) -- (1.050,0.966);
  \draw[black] (1.050,0.966) -- (1.398,0.972) -- (1.751,0.957) -- (2.101,0.965) -- (2.447,0.972) -- (2.802,0.965) -- (3.153,0.953) -- (3.499,0.956) -- (3.851,0.965) -- (4.199,0.962) -- (4.548,0.961) -- (4.899,0.959) -- (5.249,0.961) -- (5.603,0.961) -- (5.950,0.966);
  \draw[black] (5.950,0.966) -- (5.625,1.099) -- (5.295,1.238) -- (4.971,1.378) -- (4.644,1.522) -- (4.316,1.648) -- (3.992,1.794) -- (3.662,1.936) -- (3.336,2.067) -- (3.011,2.216) -- (2.682,2.338) -- (2.350,2.472) -- (2.028,2.610) -- (1.697,2.750) -- (1.374,2.883) -- (1.050,3.034);
  \draw[black] (1.050,1.173) -- (1.055,1.104) -- (1.050,1.034) -- (1.050,0.966);
  \draw[black] (1.050,0.966) -- (1.119,0.962) -- (1.187,0.962) -- (1.257,0.966);
  \draw[black] (1.050,1.173) -- (1.119,1.171) -- (1.188,1.169) -- (1.257,1.173);
  \draw[black] (1.257,1.173) -- (1.251,1.105) -- (1.258,1.035) -- (1.257,0.966);
  \draw[black] (1.415,2.880) -- (1.403,2.843) -- (1.384,2.808) -- (1.354,2.782) -- (1.336,2.746) -- (1.298,2.729) -- (1.270,2.702) -- (1.239,2.679) -- (1.200,2.669) -- (1.166,2.648) -- (1.126,2.648) -- (1.089,2.638) -- (1.050,2.629);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,0.832) {N};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,3.101) {H};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,0.832) {D};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {2,8 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {7,2 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.419,2.412) {?};
\end{tikzpicture}\end{center}


 Dans le triangle NHD rectangle en N, on cherche une relation entre l'angle aigu   \(\widehat{NHD}\) son coté adjacent et l'hypoténuse du triangle.

cos(\(\widehat{NHD}\)) = \(\dfrac{NH}{HD}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{NHD}\)) = \(\dfrac{2{,}8}{7{,}2}\)
\par\addvspace{.5em}
On a donc \(\widehat{NHD}\) =ArcCos(\(\dfrac{2{,}8}{7{,}2}\)) \(\approx\)67\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.075,3.400) -- (1.063,3.048) -- (1.082,2.697) -- (1.082,2.352) -- (1.076,1.997) -- (1.072,1.648) -- (1.085,1.297) -- (1.083,0.951) -- (1.075,0.600);
  \draw[black] (1.075,0.600) -- (1.420,0.609) -- (1.765,0.606) -- (2.111,0.593) -- (2.461,0.596) -- (2.806,0.597) -- (3.156,0.613) -- (3.503,0.611) -- (3.848,0.604) -- (4.190,0.590) -- (4.538,0.593) -- (4.887,0.593) -- (5.235,0.592) -- (5.576,0.590) -- (5.925,0.600);
  \draw[black] (5.925,0.600) -- (5.629,0.786) -- (5.321,0.961) -- (5.010,1.117) -- (4.709,1.295) -- (4.410,1.472) -- (4.104,1.653) -- (3.800,1.817) -- (3.502,2.008) -- (3.193,2.171) -- (2.890,2.345) -- (2.594,2.535) -- (2.289,2.699) -- (1.988,2.888) -- (1.680,3.052) -- (1.380,3.225) -- (1.075,3.400);
  \draw[black] (1.075,0.880) -- (1.071,0.787) -- (1.070,0.693) -- (1.075,0.600);
  \draw[black] (1.075,0.600) -- (1.168,0.604) -- (1.261,0.603) -- (1.355,0.600);
  \draw[black] (1.075,0.880) -- (1.168,0.874) -- (1.261,0.877) -- (1.355,0.880);
  \draw[black] (1.355,0.880) -- (1.362,0.786) -- (1.351,0.693) -- (1.355,0.600);
  \draw[black] (5.581,0.798) -- (5.570,0.784) -- (5.560,0.770) -- (5.556,0.753) -- (5.542,0.739) -- (5.541,0.721) -- (5.546,0.702) -- (5.530,0.688) -- (5.522,0.671) -- (5.537,0.651) -- (5.519,0.636) -- (5.531,0.617) -- (5.526,0.600);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.808,0.467) {J};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.808,3.467) {D};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.025,0.467) {V};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {4,8 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.281,0.706) {30°};
\end{tikzpicture}\end{center}


 Dans le triangle JDV rectangle en J, on cherche une relation entre l'angle aigu   \(\widehat{JVD}\) son coté adjacent et l'hypoténuse du triangle.

cos(\(\widehat{JVD}\)) = \(\dfrac{JV}{DV}\)
\par\addvspace{.5em}
  d'où    cos(30\textdegree{}) =  \(\dfrac{JV}{4{,}8}\)

On a donc JV = 4,8 \(\times\) cos(30\textdegree{}) \(\approx\) 4{,}2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,3.142) -- (1.049,2.859) -- (1.056,2.575) -- (1.056,2.287) -- (1.053,1.999) -- (1.041,1.713) -- (1.044,1.430) -- (1.050,1.142) -- (1.050,0.858);
  \draw[black] (1.050,0.858) -- (1.398,0.847) -- (1.747,0.870) -- (2.102,0.857) -- (2.452,0.867) -- (2.802,0.854) -- (3.147,0.847) -- (3.498,0.850) -- (3.849,0.868) -- (4.197,0.858) -- (4.548,0.864) -- (4.897,0.867) -- (5.248,0.870) -- (5.599,0.862) -- (5.950,0.858);
  \draw[black] (5.950,0.858) -- (5.645,1.005) -- (5.338,1.142) -- (5.034,1.288) -- (4.724,1.420) -- (4.423,1.575) -- (4.110,1.707) -- (3.806,1.859) -- (3.500,2.003) -- (3.188,2.137) -- (2.891,2.291) -- (2.580,2.426) -- (2.270,2.562) -- (1.974,2.723) -- (1.664,2.856) -- (1.362,3.009) -- (1.050,3.142);
  \draw[black] (1.050,1.086) -- (1.052,1.009) -- (1.052,0.932) -- (1.050,0.858);
  \draw[black] (1.050,0.858) -- (1.126,0.859) -- (1.203,0.860) -- (1.278,0.858);
  \draw[black] (1.050,1.086) -- (1.125,1.086) -- (1.201,1.081) -- (1.278,1.086);
  \draw[black] (1.278,1.086) -- (1.277,1.009) -- (1.279,0.934) -- (1.278,0.858);
  \draw[black] (1.412,2.974) -- (1.404,2.935) -- (1.366,2.914) -- (1.357,2.874) -- (1.326,2.850) -- (1.296,2.827) -- (1.266,2.803) -- (1.233,2.784) -- (1.200,2.765) -- (1.165,2.749) -- (1.125,2.749) -- (1.087,2.747) -- (1.050,2.745);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,0.724) {S};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,3.209) {T};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,0.724) {D};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {1,1 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.408,2.514) {65°};
\end{tikzpicture}\end{center}


 Dans le triangle STD rectangle en S, on cherche une relation entre l'angle aigu   \(\widehat{STD}\) son coté adjacent et l'hypoténuse du triangle.

cos(\(\widehat{STD}\)) = \(\dfrac{ST}{TD}\)
\par\addvspace{.5em}
d'où cos(65\textdegree{}) = \(\dfrac{ST}{1{,}1}\)

On a donc ST = 1,1 \(\times\) cos(65\textdegree{}) \(\approx\) 0,5 cm\end{document}
